[
segment tree
]
CodeForces Segment Trees, Part 1, 1.1
Problem statement
https://codeforces.com/edu/course/2/lesson/4/1/practice/contest/273169/problem/A
Solution
This segment tree will support two operations:
- Set element
i
equal tov
. - Find sum of elements in range
[r, l)
Complexity
It is O(n)
for build and O(log n)
for set
and sum
.
Code
class SegmentTree:
def __init__(self, n, arr):
self.size = 1
while self.size < n:
self.size *= 2
self.T = [0] * (2 * self.size - 1)
self.arr = arr
def _build(self, x, lx, rx):
if rx - lx == 1:
if lx < len(self.arr):
self.T[x] = self.arr[lx]
else:
mx = (lx + rx)//2
self._build(2*x+1, lx, mx)
self._build(2*x+2, mx, rx)
self.T[x] = self.T[2*x+1] + self.T[2*x+2]
def build(self):
self._build(0, 0, self.size)
def _set(self, i, v, x, lx, rx):
if rx - lx == 1:
self.T[x] = v
return
mx = (lx + rx) // 2
if i < mx:
self._set(i, v, 2 * x + 1, lx, mx)
else:
self._set(i, v, 2 * x + 2, mx, rx)
self.T[x] = self.T[2 * x + 1] + self.T[2 * x + 2]
def set(self, i, v):
self._set(i, v, 0, 0, self.size)
def _sum(self, l, r, x, lx, rx):
if l >= rx or lx >= r:
return 0
if lx >= l and rx <= r:
return self.T[x]
mx = (lx + rx) // 2
s1 = self._sum(l, r, 2 * x + 1, lx, mx)
s2 = self._sum(l, r, 2 * x + 2, mx, rx)
return s1 + s2
def sum(self, l, r):
return self._sum(l, r, 0, 0, self.size)
if __name__ == '__main__':
n, m = [int(i) for i in input().split()]
arr = [int(i) for i in input().split()]
STree = SegmentTree(n, arr)
STree.build()
for i in range(m):
t = [int(i) for i in input().split()]
if t[0] == 1:
STree.set(t[1], t[2])
else:
print(STree.sum(t[1], t[2]))